도함수의 뜻

The function is differentiable at $x=0$ if the following limit:

$$\begin{aligned} \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{f(x)-0}{x}=\lim_{x\to 0} \frac{f(x)}{x} \end{aligned}$$

exists (note that we used the fact that $f(0)=0$ as per the definition of the first branch which includes the point $x=0$). We compute left and right limits; so

$$\begin{aligned} \lim_{x\to 0^-}\frac{f(x)}{x}=\lim_{x\to 0^-}\frac{x^3-7x^2}{x}=\lim_{x\to 0^-} x^2-7x=0 \end{aligned}$$

and

$$\begin{aligned} \lim_{x\to 0^+}\frac{x^3\cos\left(\frac{1}{x}\right)}{x}=\lim_{x\to 0^+}x^2\cdot \cos\left(\frac{1}{x}\right). \end{aligned}$$

This last limit equals $0$ by the Squeeze Theorem since

$$\begin{aligned} -1\le \cos\left(\frac{1}{x}\right)\le 1 \end{aligned}$$

and so,

$$\begin{aligned} -x^2\le x^2\cdot \cos\left(\frac{1}{x}\right)\le x^2, \end{aligned}$$

where in these inequalities we used the fact that $x^2\ge 0$. Finally, since $\lim_{x\to 0^+}-x^2=\lim_{x\to 0^+}x^2=0$, the Squeeze Theorem yields that also $\lim_{x\to 0^+}x^2\cos\left(\frac{1}{x}\right) =0$, as claimed.

Since the left and right limits match (they're both equal to $0$), we conclude that indeed $f(x)$ is differentiable at $x=0$ (and its derivative at $x=0$ is actually equal to $0$).